Problem Solving

Vincen wrote: ↑

Wed Jul 22, 2020 12:46 pm

If the sequence \(a_n\) is defined as \(a_n=n^2+n+\sqrt{n+3},\) then which of the following values of \(n\) represents the first term such that \(a_n > 500?\)

A. 13
B. 22
C. 33
D. 46
E. 78

[spoiler]OA=B[/spoiler]

Source: Magoosh

So we have \(n^2+n+\sqrt{n+3} > 500\)

We have to find out the minimum positive integer value of n such that \(\sqrt{n+3}\) is an integer, and \(n^2+n+\sqrt{n+3} > 500\). By hit and trial, we find that n = 22 serves the purpose as at n = 13, though \(\sqrt{n+3}\) is an integer, \(n^2+n+\sqrt{n+3} < 500\).

Correct answer: B

Hope this helps!

-Jay
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Vincen wrote: ↑

Wed Jul 22, 2020 12:46 pm

If the sequence \(a_n\) is defined as \(a_n=n^2+n+\sqrt{n+3},\) then which of the following values of \(n\) represents the first term such that \(a_n > 500?\)

A. 13
B. 22
C. 33
D. 46
E. 78

[spoiler]OA=B[/spoiler]

Solution:

The best way to solve this problem is to try the answer choices out. Since the terms n and √(n + 3) are much smaller than n^2, we can assume a(n) = n^2.

A. If n = 13, a(13) = 13^2 = 169. This is a lot less than 500, so A can’t be the correct answer.

B. If n = 22, a(22) = 22^2 = 484. This is about 500, so let’s use the actual formula for a(n):

a(22) = 22^2 + 22 + √(22 + 3) = 484 + 22 + 5 = 511

This is greater than 500, so B is the correct answer.

Answer: B